Unfortunately my analysis lecturer, as awesome as he is, lacks the structure in his lessons to provide worked out proofs for us to use as guidelines for proving other things. Hence, I am having a great deal of trouble proving this question (or even knowing what/how to prove).

Question: Show that a sequence $\left\{ f_{n}\right\} _{n=0}^{\infty}$ of functions which converges uniformly to the function $f$ on the interval $[a,b]$ converges pointwise to $f$ on that interval.

So far... I know that a sequence of functions converges uniformly if:

For $\forall \varepsilon>0$ $\exists N$ s.t $\forall x$ and $n \geq N$, we have $|f_{n}(x)-f(x)|< \varepsilon$ (Thank you wiki)

So, in order to show pointwise convergence, I need to show $\lim_{n\rightarrow\infty}f_{n}(x)=f(x)$ for every $x$.

I can clearly see that from the definition of uniform convergence, if my $n$ is tending to infinity, that is clearly an $n \geq N$, hence if I was to take:

$|f_{n}(x)-f(x)|< \varepsilon \implies -\varepsilon<f_{n}(x)-f(x)< \varepsilon$

But $\varepsilon > 0$, so:

$f_{n}(x)-f(x)< \varepsilon \implies f_{n}(x)<f(x)+ \varepsilon$

But if I consider that, as $n \rightarrow \infty$, $\varepsilon \rightarrow 0$, hence could I say:


Hence the sequence is pointwise convergent to $f$.

Problem: I feel like I'm missing steps, or logical connections, yet I'm at a loss to fill them. I feel like my basic idea is at least close to sound, but as I said, I may just be looking at the similarities and saying "It's clearly evident", but with more work. Would greatly appreciate a critique and aide, as we don't get a great deal of feedback or particular teaching on proofs - literally, our lecturer is great, but he tends to scrawl proofs rather vaguely, leaving out much of the "self evident" or "obvious" connections.

Thank you all!

  • $\begingroup$ The definition of uniform convergence is that there is an $N$ that works for all $x$. The definition of pointwise convergence is that for any $x$ there is an $N$ which may depend on $x$. The first condition implies the second. $\endgroup$ – André Nicolas May 29 '14 at 6:25
  • $\begingroup$ Did I satisfactorily show that implication? $\endgroup$ – Yoshi May 29 '14 at 6:26
  • $\begingroup$ You may have had the right thing in mind, but it does not come through. The as $n\to\infty$, $\epsilon\to 0$ shows some confusion. In both definitions, $\epsilon$ is a fixed (though arbitrary) positive number. Really almost no "steps" are needed. $\endgroup$ – André Nicolas May 29 '14 at 6:31
  • $\begingroup$ Then how do I get an equivalence relation out of it, as I need to show for pointwise convergence? If $\varepsilon$ is fixed, then won't $f_{n}(x)$ remain strictly less than $f(x)$ for all $n$? $\endgroup$ – Yoshi May 29 '14 at 6:34
  • $\begingroup$ It need not remain less. Write down the formal definition of converges pointwise for all $x$. It reads $\forall \epsilon\gt 0 \forall x\exists N \dots$, where the stuff represented by $\dots$ is exactly the same as for uniform convergence. Then to show uniform implies pointwise is the same logic as saying if there is an $N$ who is every $x$'s friend, then every $x$ has a friend. (Note the implication does not reverse.) $\endgroup$ – André Nicolas May 29 '14 at 6:44

I suggest that you should try to write everything in such a way as to clearly show what conditions bind each statement. For example:

Uniform convergence

$f_k \to f$ uniformly on $I$ iff:

  For any real $ε > 0$:

    For some natural $n$:

      For any natural $k \ge n$:

        For any $x \in I$:

          $|f_k(x)-f(x)| < ε$

Pointwise convergence

$f_k \to f$ (pointwise) on $I$ iff:

  For any $x \in I$:

    For any real $ε > 0$:

      For some natural $n$:

        For any natural $k \ge n$:

          $|f_k(x)-f(x)| < ε$


Observe that adjacent universal quantifications can be swapped (likewise for existential quantifications), so the only actual difference between the above two is the relative position of "For some natural $n$" and "For any $x \in I$". Then it is clear that the first implies the second because if some natural $n$ works for any $x \in I$, then I can use that same particular $n$ for any $x \in I$. Here is a completely formal proof of the fact that uniform convergence implies pointwise convergence:


For any functions $f$ and $( f_k : k \in \mathbb{N} )$ such that $f_k \to f$ uniformly on $I$:

  For any $x \in I$:

    [Here we pull in the definition of uniform convergence.]

    For any real $ε > 0$:

      For some natural $n$:

        For any natural $k \ge n$:

          For any $y \in I$:

            $|f_k(y)-f(y)| < ε$

          [Here we apply the above to the $x$ in the outer scope.]

          Therefore $|f_k(x)-f(x)| < ε$

  [Now we have exactly the definition for pointwise convergence.]

  Therefore $f_k \to f$ (pointwise) on $I$


Do the same for all the other notions like "uniform continuity", "continuity", "differentiability", "Lipschitz continuity" and so on. It would really help you know what each of them really means.

  • $\begingroup$ I definitely think I'll end up doing that as part of future questions. Thanks for the massive help! It was great to see where I had not quite made the connection, and the conciseness of your definitions really helped. $\endgroup$ – Yoshi May 29 '14 at 6:59
  • $\begingroup$ @Yoshi: You're welcome! Some professors may not think quite highly of writing proofs like this, but I think those do not understand the problems that students face in understanding quantifiers. It is particularly important to fully grasp the logical structure when the quantifier complexity is high. $\endgroup$ – user21820 May 29 '14 at 7:34
  • $\begingroup$ Exactly right. Our professor does great at his explanations, but unfortunately in terms of "proof construction", I don't think I've ever seen a back-to-back proof from him all year... he generally cuts corners and omits and stack of stuff, or will do half the proof verbally, leaving you to fill in the blanks. $\endgroup$ – Yoshi May 29 '14 at 8:12

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